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Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows
 
Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows
  
 +
<math>x(t)*cos(\frac{\pi t}{4})</math> &rArr; <math>\frac{1}{2}[X(e^{j(\theta - \pi/4)})
 +
+ X(e^{j(\theta + \pi/4)}) ]</math>.<br>
  
==<math>x(t)*cos(t)</math> &rArr; <math>\frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)]</math>. ==
+
In short we are getting two side bands which look something like this
 +
 
 +
[[Image:Modulation_ECE301Fall2008mboutin.gif]]

Latest revision as of 09:24, 24 October 2008

Now we know that
$ x(t) $$ X(\omega) $

Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows

$ x(t)*cos(\frac{\pi t}{4}) $$ \frac{1}{2}[X(e^{j(\theta - \pi/4)}) + X(e^{j(\theta + \pi/4)}) ] $.

In short we are getting two side bands which look something like this

Modulation ECE301Fall2008mboutin.gif

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood